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By employing upper and lower solutions method together with maximal principle, we establish anecessary and sufficient condition for the existence of pseudo-C30, 1 as well as C20, 1 positiv

Volume 2010, Article ID 862079, 23 pages

doi:10.1155/2010/862079

Research Article

Positive Solutions for Fourth-Order Singular

p-Laplacian Differential Equations with Integral

Boundary Conditions

Xingqiu Zhang1, 2 and Yujun Cui3

1 Department of Mathematics, Huazhong University of Science and Technology, Wuhan,

Hubei 430074, China

2 Department of Mathematics, Liaocheng University, Liaocheng, Shandong 252059, China

3 Department of Applied Mathematics, Shandong University of Science and Technology,

Qingdao 266510, China

Received 7 April 2010; Accepted 12 August 2010

Academic Editor: Claudianor O Alves

Creative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited

By employing upper and lower solutions method together with maximal principle, we establish anecessary and sufficient condition for the existence of pseudo-C30, 1 as well as C20, 1 positive solutions for fourth-order singular p-Laplacian differential equations with integral boundary conditions Our nonlinearity f may be singular at t  0, t  1, and u  0 The dual results for

the other integral boundary condition are also given

1 Introduction

In this paper, we consider the existence of positive solutions for the following nonlinear

fourth-order singular p-Laplacian differential equations with integral boundary conditions:



ϕ p

xt ft, xt, xt, 0 < t < 1, x0 

0hsds < 1 and nonlinear term f satisfies the following hypothesis:

H ft, u, v : J × R × R → R is continuous, nondecreasing on u and nonincreasing

on v for each fixed t ∈ J, and there exists a real number b ∈ R such that, for any

Remark 1.2 Typical functions that satisfy condition H are those taking the form ft, u, v

 n i1 a i tu λ i m j1 b j tu −μ j , where a i , b j ∈ C0, 1, 0 < λ i < 1, μ j > 0 i  1, 2, , m;



ft, v, v, if u ≥ v ≥ 0,

v u

b

ft, v, v, if v ≥ u ≥ 0.

1.6

Boundary value problems with integral boundary conditions arise in variety ofdifferent areas of applied mathematics and physics For example, heat conduction, chemicalengineering, underground water flow, thermoelasticity, and plasma physics can be reduced

to nonlocal problems with integral boundary conditions They include two point, three point,and nonlocal boundary value problemssee 2 5 as special cases and have attracted muchattention of many researchers, such as Gallardo, Karakostas, Tsamatos, Lomtatidze, Malaguti,Yang, Zhang, and Fengsee 6 13, e.g. For more information about the general theory ofintegral equations and their relation to boundary value problems, the reader is referred to thebook by Corduneanu14 and Agarwal and O’Regan 15

Recently, Zhang et al 13 studied the existence and nonexistence of symmetricpositive solutions for the following nonlinear fourth-order boundary value problems:



ϕ p

xt ωtft, xt, 0 < t < 1, x0  x1 

ft, u, v permits singularity not only at t  0, 1 but also at v  0 By singularity, we mean that the function f is allowed to be unbounded at the points t  0, 1 and v  0.

2 Preliminaries and Several Lemmas

A function xt ∈ C20, 1 and ϕ p xt ∈ C20, 1 is called a C20, 1 positive solution of

BVP1.1 if it satisfies 1.1 xt > 0 for t ∈ 0, 1 A C20, 1 positive solution of 1.1 is

called a psuedo-C30, 1 positive solution if ϕ p xt ∈ C10, 1 xt > 0, −xt > 0 for

t ∈ 0, 1 Denote that

E  x : x ∈ C20, 1, and ϕ p

xt∈ C20, 1. 2.1

Definition 2.1 A function αt ∈ E is called a lower solution of BVP 1.1 if αt satisfies



ϕ p

αt≤ ft, αt, αt, 0 < t < 1, α0 −

To prove the main results, we need the following maximum principle

Lemma 2.3 Maximum principle If x ∈ F k , such that ϕ p xt ≥ 0, t ∈ a k , b k , then xt ≥ 0, −xt ≥ 0, t ∈ a k , b k .

zt  za k  za k t − a k −

t

a t − sσsds. 2.15

Let t  b kin2.15, we obtain that

Substituting2.20 into 2.17, we have

Obviously, G k t, s ≥ 0, K k t, s ≥ 0, σ 2k ≥ 0 From 2.21, it is easily seen that zt ≥ 0 for

t ∈ a k , b k  By 2.11, we know that ϕ p yt ≥ 0, that is, yt ≥ 0 Thus, we have proved that

−xt ≥ 0, t ∈ a k , b k Similarly, the solution of 2.5 and 2.7 can be expressed by

Lemma 2.4 Suppose that H holds Let xt be a C20, 1 positive solution of BVP 1.1 Then there exist two constants 0 < I1< I2such that

It is easy to see that

Lemma 2.5 Suppose that H holds And assume that there exist lower and upper solutions of BVP

1.1, respectively, αt and βt, such that αt, βt ∈ E, 0 ≤ αt ≤ βt for t ∈ 0, 1, α1  β1  0 Then BVP 1.1 has at least one C20, 1 positive solution xt such that αt ≤ xt ≤ βt,

t ∈ 0, 1 If, in addition, there exists Ft ∈ L10, 1 such that

ft,x,x ≤ Ft, for αt ≤ xt ≤ βt, 2.32

then the solution xt of BVP 1.1 is a pseudo-C30, 1 positive solution.

Proof For each k, for all xt ∈ E k  {x : x ∈ C2a k , b k , and ϕ p xt ∈ C2a k , b k}, wedefined an auxiliary function

f

t, βt, βt, if xt ≥ βt.

2.33

By conditionH, we have that F k : E k → 0, ∞ is continuous.

Let{a k }, {b k } be sequences satisfying 0 < · · · < a k 1 < a k < · · · < a1 < b1 < · · · < b k <

b k 1 < · · · < 1, a k → 0 and b k → 1 as k → ∞, and let {r ki }, i  1, 2, 3, be sequences satisfying

For convenience, we define linear operators as follows:

compact operator and A k ϕ−1

p B k E k  is a relatively compact set So, A k ϕ−1

p B k F k  : E k →

E k is a completely continuous operator In addition, x ∈ E k is a solution of2.35 if and

only if x is a fixed point of operator A k ϕ−1

p B k F k x  x Using the Shauder’s fixed point theorem, we assert that A k ϕ−1

p B k F k  has at least one fixed point x k ∈ C2a k , b k , by x k t 

A k ϕ−1

p B k F k x k t, we can get ϕ p x k ∈ C2a k , b k .

We claim that

αt ≤ x k t ≤ βt, t ∈ a k , b k . 2.38From this it follows that



ϕ p

x

k t ft, βt, βt, t ∈ a k , b k . 2.41

On the other hand, since βt is an upper solution of 1.1, we also have



ϕ p

βt≥ ft, βt, βt, t ∈ a k , b k . 2.42Then setting

−βt − x

k t≥ 0, t ∈ a k , b k . 2.46Set

ut  βt − x k t, t ∈ a k , b k . 2.47Then

−ut ≥ 0, t ∈ a k , b k , x ∈ C2a k , b k , ua k −

In addition, if2.32 holds, then |ϕ p xt| ≤ Ft Hence, ϕ p xtis absolutelyintegrable on0, 1 This implies that xt is a pseudo-C30, 1 positive solution of 1.1

3 The Main Results

Theorem 3.1 Suppose that H holds, then a necessary and sufficient condition for BVP 1.1  to have a pseudo-C30, 1 positive solution is that the following integral condition holds:

0 <

1

0fs, 1 − s, 1 − sds < ∞. 3.1

Proof The proof is divided into two parts, necessity and suffeciency.

Necessity Suppose that xt is a pseudo-C30, 1 positive solution of 1.1 Then both

It follows from3.3 and 3.6 that

0 <

1

0fs, 1 − s, 1 − sds < ∞, 3.7which is the desired inequality

Sufficiency First, we prove the existence of a pair of upper and lower solutions Since ξl/l2

It is easy to know from3.11 and 3.12 that ϕ p −bt ∈ C10, 1 ByLemma 2.4, we know

that there exists a positive number k < 1 such that

l1k1



−l1

1 ≤ 0. 3.15

αt  l1b1t, βt  l1

1

b1t, t ∈ 0, 1. 3.16Thus, from3.14 and 3.16, we have

l1k1



l1k1−b ft, 1 − t, 1 − t  Ft.

3.20

From3.1, we have1

0Ftdt < ∞ So, it follows fromLemma 2.5that BVP1.1 admits a

pseudo-C30, 1 positive solution such that αt ≤ xt ≤ βt.

Remark 3.2 Lin et al.23,24 considered the existence and uniqueness of solutions for somefourth-order andk, n − k conjugate boundary value problems when ft, u, v  qtgu hv, where

g : 0, ∞ −→ 0, ∞ is continuous and nondecreasing,

h : 0, ∞ −→ 0, ∞ is continuous and nonincreasing, 3.21

under the following condition:

P1 for t ∈ 0, 1 and u, v > 0, there exists α ∈ 0, 1 such that

Obviously, 3.21-3.22 imply condition P2 and condition P2 implies condition

H So, condition H is weaker than conditions P1 and P2 Thus, functions considered inthis paper are wider than those in23–26

In the following, when ft, u, u admits the form ft, u, that is, nonlinear term f is not mixed monotone on u, but monotone with respect u, BVP1.1 becomes



ϕ p

xt ft, xt, 0 < t < 1, x0 

If f t, u satisfies one of the following:

H∗ ft, u : J × R → R is continuous, nondecreasing on u, for each fixed t ∈ 0, 1, there exists a function ξ : 1, ∞ → R , ξl < l and ξl/l2is integrable on1, ∞

such that

ft, lu ≤ ξlft, u, ∀t, u ∈ J × R , l ∈ 1, ∞. 3.24

Theorem 3.3 Suppose that H∗ holds, then a necessary and sufficient condition for BVP 3.23 to have a pseudo-C30, 1 positive solution is that the following integral condition holds

0 <

1

0fs, 1 − sds < ∞. 3.25

Proof The proof is similar to that ofTheorem 3.1; we omit the details

Theorem 3.4 Suppose that H∗ holds, then a necessary and sufficient condition for problem 3.23

to have a C20, 1 positive solution is that the following integral condition holds

0 <

1

0

s1 − sfs, 1 − sds < ∞. 3.26

Proof The proof is divided into two parts, necessity and suffeciency.

Necessity Assume that xt is a C20, 1 positive solution of BVP 3.23 ByLemma 2.4, there

exist two constants I1and I2, 0 < I1< I2, such that

Sufficiency Suppose that 3.26 holds Let

Thus,3.12, 3.39, and 3.40 imply that 0 ≤ b2t < ∞ ByLemma 2.4, we know that there

exists a positive number k2< 1 such that

l2k2



− 1

l2 ≤ 0. 3.42Let

αt  l2b2t, βt  1

l2

b2t, t ∈ 0, 1. 3.43Thus, from3.41 and 3.43, we have

l2



ft, 1 − t ≤

1

From3.39 and 3.43, it follows that

From the first conclusion ofLemma 2.5, we conclude that problem1.1 has at least

one C20, 1 positive solution xt.

By analogous methods, we have the following results.

Assume that xt is a C20, 1 positive solution of problem 4.1 Then xt can be

If xt is a C20, 1 positive solution of problem 4.8 Then xt can be expressed by

If xt is a C20, 1 positive solution of problem 4.15 Then xt can be expressed by

xt  A1ϕ−1

p B2ft, xt, xt. 4.18

Theorem 4.7 Suppose that H holds, then a necessary and sufficient condition for problem 4.15

to have a pseudo-C30, 1 positive solution is that the following integral condition holds:

0 <

1

0fs, s, sds < ∞. 4.19

Theorem 4.8 Suppose that H∗ holds, then a necessary and sufficient condition for problem 4.16

to have a pseudo-C30, 1 positive solution is that the following integral condition holds:

0 <

1

0

fs, sds < ∞. 4.20

Theorem 4.9 Suppose that H∗ holds, then a necessary and sufficient condition for problem 4.16

to have a C20, 1 positive solution is that the following integral condition holds:

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